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In non-relativistic physics, momentum defined as $p = mv $, so as object velocity increased, then the momentum increased. Or if mass of object increased/decreased, then the momentum too increased or decreased (as in rocket). And this concept very useful when discussing elastec collision between two object to determine how the state of both object after collision.

But in relativistic theory, momentum defined as $$ p = \frac{ m v}{\sqrt{1 - \frac{v^2}{c^2 }} } .$$ In this formulation, $m$ keeps constant, because there is no relativistics mass. So classical approximation for this variable is $$ \frac{ m v}{\sqrt{1 - \frac{v^2}{c^2 }} } \approx m v$$ or $$v_\text{rel} = \frac{ v}{\sqrt{1 - \frac{v^2}{c^2 }} } \approx v ,$$ because $m$ not changed as Einstein say.

So, what is physical meaning of this $v_\text{rel} $. Is it imply that $\textbf{ velocity }$ of object changed when its $\textbf{velocity}$ approach the velocity of light?

Edit: And what constitute $\gamma v $ for the planet orbiting the sun, if the mass stay constant?

Qmechanic
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Mohammad Fajar
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  • Relativistic mass is an outdated concept. See https://physics.stackexchange.com/q/133376/ . What makes you think that the quantity $v_\text{rel}$ you define has any special physical interpretation? I don't think it does. The title of the question doesn't seem to relate to what is asked in the question. –  Sep 10 '17 at 19:38
  • because we need to measure it. – Mohammad Fajar Sep 10 '17 at 19:41
  • @BenCrowell it does actually have a physical interpretation, it's the spatial component of the four-velocity. – Javier Sep 10 '17 at 19:44
  • it is impossible to do physics without measurement. For classical momentum, we can measure it by measure mass and velocity and multiply. But for relativistics formulation there is no clear definition. Is it just a methamatical formula without physical correspondence? Or there is some hidden variable in relativistic momentum that needless to say, why there is no even a single book that mention it? – Mohammad Fajar Sep 10 '17 at 19:45
  • yes, but what I needed is bottom-up approach. We discover relativistics momentum before formulated it into 4-velocities. Chronologically, it is impossible to start physics discovery from 4-velocities and goes into relativistics momentum, because doing physics imply we must measure something, or I'am wrong? – Mohammad Fajar Sep 10 '17 at 19:48
  • I'm not sure that you really have as clear an understanding of classical momentum as you think you do. Just saying 'oh we multiply these things' and treating it as better understood than 'we plug those same two things into a slightly more complicated formula' does not strike me as a difference of type, just one of degree. But if you understood what momentum means in some sense that agrees with the classical mathematics then you can ask if it also agrees with the relativistic mathematics (which it does as I indicate below). – dmckee --- ex-moderator kitten Sep 10 '17 at 20:43

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Momentum puts a numeric value on inertia in the sense that it represents how much impulse is needed to stop an object (or conversely was needed to get it up to speed from rest in the first place).

For a constant force, impulse is the applied force times the duration of application (i.e. how hard and long you have to push), and because of Newton's second law we see that a net impulse of $J = F \Delta t$ applied to an object of mass $m$ can accelerate it from rest to speed $$v = a\, \Delta t = \frac{F}{m} \Delta t \;. $$ Of course such acceleration is a gradual process, shortly after the initial application of the force the object has a small speed, and the next short period of application adds a little more speed and so on.

The only thing that is different about the relativisitic case is the way in which velocity adds. Adding velocity $u$ to an aboject already going at $v$ doesn't result in a total of $V_\text{classical} = u + v$ as appears to be the case at low relative speeds, but in $$ V_\text{relativisitc} = \frac{u + v}{1 + \frac{uv}{c^2}} .$$ If you apply the correct (relativistic) velocity composition rule to the accelerating object you end with it's velocity asymptotically approaching $c$ rather than growing without bound, but the impulse needed to get it to those speeds continues to be $Ft$ and does increase without bound, so $$ p = \frac{mv}{\sqrt{1 - (v/c)^2}} $$ as you said.

To re-iterate the fact I started with momentum measures the impulse needed to stop an object, and that interpretation is completely consistent in either classical or relativistic physics.

Aside: we measure the relativistic momentum of particles in accelerators all the time (as in it's a work-a-day process) using their bending radius in magnetic fields. Einstein's theory passes this test with flying colors.

dmckee --- ex-moderator kitten
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  • there is two distinction in this illustration, object A that give an action (or force) it is magnetic accelerator, and object B that given an action (electron or something). For object A we can use F dt as a measurable variable for its momentum definition. But for object B, are the same apply too? In fact as outside oberver we can only measure $m$ and $v$ for object B. So, based formula above, $p = \gamma m v$, then if object A give $F dt $, what we expect occur in object B is increasing in its velocity according to formula $v_\textbf{rel} = \gamma v$ – Mohammad Fajar Sep 10 '17 at 21:19
  • What i say was, we can measure a force ($F dt $) given by accelerator, but for electron, we must measure $ \gamma v $. And I dont know what is interpretation of $\gamma v$? – Mohammad Fajar Sep 10 '17 at 21:24
  • The Lorentz factor ($\gamma$) only depends on $v$. Knowing $m$ and $v$ is enough. I suspect that you are getting really hung up on trying to say 'I want to break this formula into these* pieces and interpret the pieces'*, when the proper way to understand the system is in a different decomposition. There is no need and no reason to find an interpretation for $\gamma v$, but if you really want one it is the speed that the object would have had if subjected to the same forces in a Galilean rather than Einsteinian universe—a pointless and unhelpful subjunctive. – dmckee --- ex-moderator kitten Sep 10 '17 at 21:39
  • Ok, this is the last comment: so the energy of a single electron to make a round trajectory (with velocity near the speed oflight) is equal with energy expend by the accelerator to move that electron? And this energy consist of energy of power source that drift by movement of many of electron to make that accelerator works?

    And this can generalized to momentum variable, so momentum of that single electron, just the total momentum/energy produced by accelerator?

     – Mohammad Fajar Sep 10 '17 at 21:52
  • You should distinguish clearly in your head between energy and momentum. But accelerators are highly inefficicent machines: they require much more energy to get particles up to speed than the kinetic energy the particles gain in the process. – dmckee --- ex-moderator kitten Sep 10 '17 at 22:01
  • to make thing clear I ask philosophical question to make this concept useful not just for electron, but also for other things in universe. What if we travel in a spaceship with velocity near speed of light and try to measure/determine its momentum? So when we get this $p$, its value approximate (just a little different) with classical definition $p = m v$ for slower velocity? – Mohammad Fajar Sep 11 '17 at 04:46
  • Or for macroscopic object (like planet orbiting the sun) what variable that reside in $\gamma v$ so we can measure the difference of the momentum for different value if $v$?. So there is the difference for its momentum when he move with $v_1$ and with $v_2$. And this difference not linearly related to $m v$, there is a little adds? So when the $v$ become smaller, we just use classical definition $ p = mv$? – Mohammad Fajar Sep 11 '17 at 04:50
  • What constitute $\gamma v$ for the planet orbiting the sun, if the mass stay constant? – Mohammad Fajar Sep 11 '17 at 04:56
  • Or I can say that the time dilation formula, $t = \frac{t_0}{\sqrt{1 - v^2 / c^2}}$ have no any physical meaning, it is just to make sure that "everything" becomes invariant under Lorentz Transformation? – Mohammad Fajar Sep 11 '17 at 07:04